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Math Help - Cylindrical coordinates to cartesian coordinates

  1. #1
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    Cylindrical coordinates to cartesian coordinates

    Hi there. Hi have in cylindrical coordinates that \theta=\displaystyle\frac{\pi}{3}, and I must make the graph, and take it into cartesian coordinates. How should I do?

    I've tried this way:

    \begin{Bmatrix}x=r\cos\displaystyle\frac{\pi}{3}\\  y=r\sin\displaystyle\frac{\pi}{3} \\z=z\end{matrix}\Rightarrow{\begin{Bmatrix}x=\dis  playstyle\frac{r}{2}\\y={r\displaystyle\frac{\sqrt[ ]{3}}{2} \\z=z\end{matrix}}

    I think its a semi-plane parallel to the line: 2\displaystyle\frac{y}{\sqrt[ ]{3}}-2x=0. I thought of working geometrically with it, taking another point. Or taking three points, but I think its probably easier someway, just from the equations system. I don't know how to take x and y, to make them a function of z.

    Bye there!
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Ulysses View Post
    Hi there. Hi have in cylindrical coordinates that \theta=\displaystyle\frac{\pi}{3}, and I must make the graph, and take it into cartesian coordinates. How should I do?

    I've tried this way:

    \begin{Bmatrix}x=r\cos\displaystyle\frac{\pi}{3}\\  y=r\sin\displaystyle\frac{\pi}{3} \\z=z\end{matrix}\Rightarrow{\begin{Bmatrix}x=\dis  playstyle\frac{r}{2}\\y={r\displaystyle\frac{\sqrt[ ]{3}}{2} \\z=z\end{matrix}}

    I think its a semi-plane parallel to the line: 2\displaystyle\frac{y}{\sqrt[ ]{3}}-2x=0. I thought of working geometrically with it, taking another point. Or taking three points, but I think its probably easier someway, just from the equations system. I don't know how to take x and y, to make them a function of z.

    Bye there!
     \theta = \frac{ \pi }{3} corresponds to special triangles. Special right triangles - Wikipedia, the free encyclopedia
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  3. #3
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    How do I apply that on this problem?

    Thanks btw.
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  4. #4
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    Hello, Ulysses!

    Your work is correct . . . just put it together.


    \text{In cylindrical coordinates: }\;\theta\:=\: \frac{\pi}{3}

    \text{I must take it into cartesian coordinates and graph.}


    \text{I've tried this way:}

    \begin{Bmatrix}x&=&r\cos\frac{\pi}{3}\\ \\[-3mm] y&=&r\sin\frac{\pi}{3} \\ \\[-3mm] z&=&z\end{matrix}\quad\Rightarrow\quad \begin{Bmatrix}x&=& \frac{1}{2}r & [1] \\ \\[-3mm] y&=&{\frac{\sqrt{3}}{2}\,\!r & [2] \\ \\[-3mm] z&=&z\end{matrix}}


    \text{I think its a semi-plane parallel to the line: }\:\frac{2}{\sqrt{3}}y-2x\:=\:0
    Um, it's the entire plane . . .

    \begin{array}{cccccc}\text{From [1], we have: }& r &=& 2x \\ \\[-3mm]<br />
\text{From [2]. we have:} & r &=& \frac{2}{\sqrt{3}}y \end{array}

    \text{Hence: }\;\frac{2}{\sqrt{3}}y \:=\:2x \quad\Rightarrow\quad y \:=\:\sqrt{3}\,x


    \text{This is a vertical plane passing through the }z\text{-axis,}
    . . \text{making a }60^o\text{ angle with the positive }x\text{-axis.}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    By the way . . .

    . . \frac{2}{3} produces: . \frac{2}{3}

    . . \dfrac{2}{3} produces: . \dfrac{2}{3}
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  5. #5
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    Thank you very much Soroban.

    So, I must use the fact of that its a plane parallel to that line, and work geometrically to get it into a function of z, right?
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  6. #6
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    It'll be a semi-plane if you're prohibiting negative r-values, and a full plane if you're allowing negative r-values. Polar coordinates usually allow negative r-values. By extension, then, I would think cylindrical coordinates would usually allow negative r-values.
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  7. #7
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, Ulysses!

    Your work is correct . . . just put it together.



    \begin{array}{cccccc}\text{From [1], we have: }& r &=& 2x \\ \\[-3mm]<br />
\text{From [2]. we have:} & r &=& \frac{2}{\sqrt{3}}y \end{array}

    \text{Hence: }\;\frac{2}{\sqrt{3}}y \:=\:2x \quad\Rightarrow\quad y \:=\:\sqrt{3}\,x


    \text{This is a vertical plane passing through the }z\text{-axis,}
    . . \text{making a }60^o\text{ angle with the positive }x\text{-axis.}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    By the way . . .

    . . \frac{2}{3} produces: . \frac{2}{3}

    . . \dfrac{2}{3} produces: . \dfrac{2}{3}
    Note how this corresponds to the special triangles noted above.

    The angle  \frac{ \pi }{3} is given by the triangle with the sides  \sqrt{3} , and  1 with the hypotenuse  2 .

    Rise over run gives  \frac{y}{x} = \frac{ \sqrt{3} }{1} \to y = \sqrt{3}x
    This does not have a z component.
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